We’re now going to move to our final rule.
This is called the quotient rule and we use this when you have two functions dividing each other.
So, we’ve previously looked at the chain rule when we look at the function of a function.
We’ve looked at the product rule when you look at two functions multiplying together
and now we’re looking at two functions dividing each other
and for that we’re going to introduce the quotient rule.
If you observe this question or this example that we have in front of us,
we now have 5x to the 4 that is over x plus 3 all squared.
This is the first time we’re looking at a function like this
because we have dealt with functions within functions
and we have dealt with functions that are timesing each other
and we?ve dealt with functions that are fairly easier to differentiate.
But here we are now faced with a function that is being divided by a function.
So we’re going to introduce our new rule.
We're going to give you the definition of the quotient rule.
We’ll do a numerical example using the definition
and then we?ll also do a quick smaller proof this time,
much smaller than chain and product rule to prove that the actual rule is true.
Firstly the definition, if you’re given a function f of x divided by g of x, f of x and g of x are any two functions
any two expressions, algebraic expressions that are dividing each other.
The gradient is given by the differential of f of x, so that?s a differential of the function at the top.
And you leave g of x as it is, so the function at the bottom stays as it is.
You then subtract, so notice the differences between this and the product rule.
This time there is a minus in between.
You then subtract f of x but stays as it is and then you multiply it with the differential of g of x.
You divide it all by g(x) squared and you can write this much faster as we did with the product rule.
If you want to call f of x, u and you call g of x, v or the function at the top you call u
and the function at the bottom you call v, it’s much easier perhaps to learn the second rule
which just states that vdudx minus udvdx divided by v squared
will give you the gradient of two functions dividing each other.
So make sure you learn this and also make sure that you know the difference between this and the product rule.
They’re two very similar looking rule so it’s easy to get confused
but remember that the minus sign is a positive sign when you do product rule and it’s minus in this case.
And then in the quotient rule you also divide it by a term v squared
whereas in the product rule you don?t have to divide it by anything.
Let’s do a numerical example and apply this rule now.
So we look at the question, our example here which is asking you to differentiate y equals to 5x to the 4,
x plus 3 to the power of 2. First, you notice that this is a quotient rule question
because you have a function at the top which you call u and the function at the bottom that you call v.
There's very little room to be creative with the quotient rule,
so remember always to call your upper function u and your lower function v
because that’s what the rule is based around. So don’t try and mix it up and call the top function v
or the bottom function u because then the rule won’t work.
Remember what we said that the definition was so we’re saying that dy/dx is going to be vdudx
minus udvdx all over v squared. You can write this in terms of dy/dx if you prefer to
so you could write this as vdu/dx minus udv/dx, all over v squared.
And you can also write it in terms of the functions that we showed earlier on the slide.
So, let’s try to apply this rule. We're going to take our top function u which is 5x to the 4,
our bottom function v which is x plus 3 all squared.
We're going to differentiate each one of them separately so we’ll do du/dx here which gives us 20x to the 3.
Remember what we?ve done, bring the power down and decrease the power by 1.
We are going to dv/dx so we're going to differentiate v.
Observe now, we're faced with another chain rule. So we have a big function
and a little function on the inside. For the chain rule, you differentiate the big function first
and then multiply it with the differential of the little function.
So we bring 2 to the front, anything that’s on the inside stays as is to the power of 1.
Multiply with the differential of the inside so when you differentiate x,
will just go to 1 and the 3 just disappears, so nice and straightforward.
So we’ve got 2, x plus 3 or you can times it through to give you 2x plus 6.
Okay, now we put these all into the quotient rule so we can now say that dy/dx is vdudx.
So we’re going to start with this v and we’re going to multiply it with du/dx
and then we'll say our u function and we’re going to multiply with dv/dx but the order is really important.
You must start with the v so that you don’t get it wrong because of the minus in between.
So I?m going to multiply v with du/dx which gives me 20x cubed multiply it by x plus 3 squared.
Remember to put a minus in between and I’m now going to multiply this term
with this term here giving me 5x to the 4 multiply it with 2x plus 6
and we have that all over v squared. So this is my v term here,
we now want to square it so we’ll have x plus 3 squared but then you also square it again
because it’s v squared. Have a little look if we can tidy this up,
there’s not much point expanding this bracket so we’ll just leave this as it is so we can do 20x cubed x plus 3 squared.
You could times this through but it’s already factorized so it?s probably better for us to leave it in its factorized form.
If you can see things that might simplify then it’s a good idea to expand it
and then we can simplify it but if it doesn’t it’s better to just leave it in a factorized version.
When you have a power raised by another power, so this, remember what you do
in this case you’re going to multiply the powers so at the bottom you’ll be left with x plus 3 to the power of 4.
And again this answer would get fairly messier if we started to expand all of these brackets
and brought them together, so we’re okay to just leave it like it is because it’s all factorized.
And this now is our derivative or the gradient of this function here
so if we are differentiating two functions that are dividing each other,
we use the quotient rule and we get to a fairly complicated derivative.
And then remember what you can do with this,
you can substitute numbers into here to find the gradient at particular points.
From there you can find equations of tangents and normals and lots of other information about this graph.